Given:
The equation for the linearized regression line is:
[tex]\log y=0.30x+0.296[/tex]
where x represents number of weeks and y be the customer's stock.
To find:
The number of weeks that will pass before the value of the stock reaches $200.
Solution:
We have,
[tex]\log y=0.30x+0.296[/tex]
Substituting y=200, we get
[tex]\log (200)=0.30x+0.296[/tex]
[tex]2.301=0.30x+0.296[/tex]
[tex]2.301-0.296=0.30x[/tex]
[tex]2.005=0.30x[/tex]
Divide both sides by 0.30.
[tex]\dfrac{2.005}{0.30}=x[/tex]
[tex]6.6833333=x[/tex]
[tex]x\approx 6.7[/tex]
Therefore, the correct option is C.
